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# Unitarily invariant norm inequalities involving Heron and Heinz means

Journal of Inequalities and Applications20142014:288

https://doi.org/10.1186/1029-242X-2014-288

• Received: 16 April 2014
• Accepted: 4 July 2014
• Published:

## Abstract

In this paper, we present some new inequalities for unitarily invariant norms involving Heron and Heinz means for matrices, which generalize the result of Theorem 2.1 (Fu and He in J. Math. Inequal. 7(4):727-737, 2013) and refine the inequality of Theorem 6 (Zhan in SIAM J. Matrix Anal. Appl. 20: 466-470, 1998). Our results are a refinement and a generalization of some existing inequalities.

MSC:47A30, 15A60.

## Keywords

• unitarily invariant norm
• positive definite matrices
• convex function
• Heron means
• Heinz means

## 1 Introduction

Throughout, let ${M}_{m,n}$ be the space of $m×n$ complex matrices and ${M}_{n}={M}_{n,n}$.

A norm $\parallel \cdot \parallel$ is called unitarily invariant norm if $\parallel UAV\parallel =\parallel A\parallel$ for all $A\in {M}_{n}$ and for all unitary matrices $U,V\in {M}_{n}$. Two classes of unitarily invariant norms are especially important. The first is the class of the Ky Fan k-norm ${\parallel \cdot \parallel }_{\left(k\right)}$, defined as
${\parallel A\parallel }_{\left(k\right)}=\sum _{j=1}^{k}{s}_{j}\left(A\right),\phantom{\rule{1em}{0ex}}k=1,\dots ,n,$
where ${s}_{i}\left(A\right)$ ($i=1,\dots ,n$) are the singular values of A with ${s}_{1}\left(A\right)\ge \cdots \ge {s}_{n}\left(A\right)$, which are the eigenvalues of the positive semidefinite matrix $|A|={\left({A}^{\ast }A\right)}^{\frac{1}{2}}$, arranged in decreasing order and repeated according to multiplicity. The second is the class of the Schatten p-norm ${\parallel \cdot \parallel }_{\left(p\right)}$, defined as
${\parallel A\parallel }_{p}={\left(\sum _{j=1}^{n}{s}_{j}^{p}\left(A\right)\right)}^{\frac{1}{p}}={\left(tr{|A|}^{p}\right)}^{\frac{1}{p}},\phantom{\rule{1em}{0ex}}1\le p<\mathrm{\infty }.$
For two nonnegative real numbers a and b, the Heinz mean and Heron mean in the parameter v, $0\le v\le 1$, are defined, respectively, as
$\begin{array}{c}{H}_{v}\left(a,b\right)=\frac{{a}^{v}{b}^{1-v}+{a}^{1-v}{b}^{v}}{2},\hfill \\ {F}_{\alpha }\left(a,b\right)=\left(1-\alpha \right)\sqrt{ab}+\alpha \frac{a+b}{2}.\hfill \end{array}$

Note that ${H}_{0}\left(a,b\right)={H}_{1}\left(a,b\right)=\frac{a+b}{2}$ (the arithmetic mean of a and b) and ${H}_{\frac{1}{2}}\left(a,b\right)=\sqrt{ab}$ (the geometric mean of a and b). It is easy to see that as a function of v, ${H}_{v}\left(a,b\right)$ is convex, attains its minimum at $v=\frac{1}{2}$, and attains its maximum at $v=0$ and $v=1$.

The operator version of the Heinz mean  asserts that if A, B and X are operators on a complex separable Hilbert space such that A and B are positive, then for every unitarily invariant norm $\parallel \cdot \parallel$, the function $g\left(v\right)=\parallel {A}^{v}X{B}^{1-v}+{A}^{1-v}X{B}^{v}\parallel$ is convex on $\left[0,1\right]$, attains its minimum at $v=\frac{1}{2}$, and attains its maximum at $v=0$ and $v=1$. Moreover, the operator version of the Heron mean  asserts that $f\left(\alpha \right)=\parallel \left(1-\alpha \right){A}^{\frac{1}{2}}X{B}^{\frac{1}{2}}+\alpha \left(\frac{AX+XB}{2}\right)\parallel$.

Let $A,B,X\in {M}_{n}$, A, B are positive definite, Kaur and Singh  have proved the following inequalities for any unitarily invariant norm $\parallel \cdot \parallel$:
$\frac{1}{2}\parallel {A}^{v}X{B}^{1-v}+{A}^{1-v}X{B}^{v}\parallel \le \parallel \left(1-\alpha \right){A}^{\frac{1}{2}}X{B}^{\frac{1}{2}}+\alpha \left(\frac{AX+XB}{2}\right)\parallel$
(1.1)
and
$\parallel {A}^{\frac{1}{2}}X{B}^{\frac{1}{2}}\parallel \le \frac{1}{2}\parallel {A}^{\frac{2}{3}}X{B}^{\frac{1}{3}}+{A}^{\frac{1}{3}}X{B}^{\frac{2}{3}}\parallel \le \frac{1}{2+t}\parallel AX+t{A}^{\frac{1}{2}}X{B}^{\frac{1}{2}}+XB\parallel ,$
(1.2)

where $\frac{1}{4}\le v\le \frac{3}{4}$, $\alpha \in \left[\frac{1}{2},\mathrm{\infty }\right)$ and $t\in \left(-2,2\right]$.

Replacing A, B by ${A}^{2}$, ${B}^{2}$ in (1.1) and (1.2), then putting $u=2v$, the following inequalities hold:
$\frac{1}{2}\parallel {A}^{u}X{B}^{2-u}+{A}^{2-u}X{B}^{u}\parallel \le \parallel \left(1-\alpha \right)AXB+\alpha \left(\frac{{A}^{2}X+X{B}^{2}}{2}\right)\parallel$
(1.3)
and
$\parallel AXB\parallel \le \frac{1}{2}\parallel {A}^{\frac{4}{3}}X{B}^{\frac{2}{3}}+{A}^{\frac{2}{3}}X{B}^{\frac{4}{3}}\parallel \le \frac{1}{t+2}\parallel {A}^{2}X+tAXB+X{B}^{2}\parallel ,$
(1.4)

where $\frac{1}{2}\le u\le \frac{3}{2}$, $\alpha \in \left[\frac{1}{2},\mathrm{\infty }\right)$ and $t\in \left(-2,2\right]$.

Zhan proved in  that if $A,B,X\in {M}_{n}$, such that A, B are positive semidefinite, then
$\parallel {A}^{u}X{B}^{2-u}+{A}^{2-u}X{B}^{u}\parallel \le \frac{2}{t+2}\parallel {A}^{2}X+tAXB+X{B}^{2}\parallel$
(1.5)

for $\frac{1}{2}\le u\le \frac{3}{2}$ and $t\in \left(-2,2\right]$.

Let $A,B,X\in {M}_{n}$, such that A, B are positive semidefinite, for $\frac{1}{2}\le u\le \frac{3}{2}$ and $t\in \left(-2,2\right]$; Fu et al. in  proved that
$\begin{array}{c}2\parallel AXB\parallel +2\left({\int }_{\frac{1}{2}}^{\frac{3}{2}}\parallel {A}^{r}X{B}^{2-r}+{A}^{2-r}X{B}^{r}\parallel \phantom{\rule{0.2em}{0ex}}dr-2\parallel AXB\parallel \right)\hfill \\ \phantom{\rule{1em}{0ex}}\le \frac{2}{t+2}\parallel {A}^{2}X+tAXB+X{B}^{2}\parallel .\hfill \end{array}$
(1.6)

Recently, Kaur et al. , He et al.  and Bakherad et al.  have studied similar topics.

For the sake of convenience, we set
$g\left(u\right)=\parallel \frac{{A}^{u}X{B}^{2-u}+{A}^{2-u}X{B}^{u}}{2}\parallel .$

In Section 2, we will generalize and refine some existing inequalities for unitarily invariant norms involving Heron and Heinz means for matrices and present some new refinements of the inequalities above.

## 2 Main results

In this section, we firstly utilize the convexity of the function $g\left(u\right)$ to obtain a unitarily invariant norms inequality that leads to another version of the inequality (1.6), which is also the refinement of the inequality (1.5).

To obtain the results, we need the following lemma on convex functions [8, 9].

Lemma 2.1 Let f be a real valued continuous convex function on an interval $\left[a,b\right]$ which contains $\left({x}_{1},{x}_{2}\right)$. Then for ${x}_{1}\le x\le {x}_{2}$, we have
$f\left(x\right)\le \frac{f\left({x}_{2}\right)-f\left({x}_{1}\right)}{{x}_{2}-{x}_{1}}x-\frac{{x}_{1}f\left({x}_{2}\right)-{x}_{2}f\left({x}_{1}\right)}{{x}_{2}-{x}_{1}}.$
Theorem 2.2 Let $A,B,X\in {M}_{n}$, such that A, B are positive semidefinite. Then for any unitarily invariant norm $\parallel \cdot \parallel$, $\frac{1}{2}\le u\le \frac{3}{2}$ and $\alpha \in \left[\frac{1}{2},\mathrm{\infty }\right)$,
$\begin{array}{rcl}\parallel {A}^{u}X{B}^{2-u}+{A}^{2-u}X{B}^{u}\parallel & \le & 2\left(4{r}_{0}-1\right)\parallel AXB\parallel \\ +4\left(1-2{r}_{0}\right)\parallel \left(1-\alpha \right)AXB+\alpha \left(\frac{{A}^{2}X+X{B}^{2}}{2}\right)\parallel ,\end{array}$
(2.1)

where ${r}_{0}=min\left[\frac{u}{2},1-\frac{u}{2}\right]$.

Proof For $\frac{1}{2}\le u\le 1$, by the convexity of the function $g\left(u\right)$ and Lemma 2.1, presented above, we have
$g\left(u\right)\le \frac{g\left(1\right)-g\left(\frac{1}{2}\right)}{\frac{1}{2}}u-\frac{\frac{1}{2}g\left(1\right)-g\left(\frac{1}{2}\right)\right)}{\frac{1}{2}},$
which implies
$g\left(u\right)\le 2\left(1-u\right)g\left(\frac{1}{2}\right)+\left(2u-1\right)g\left(1\right).$
(2.2)
By (1.3) and (2.2), we have
$\parallel {A}^{u}X{B}^{2-u}+{A}^{2-u}X{B}^{u}\parallel \le 4\left(1-u\right)\parallel \left(1-\alpha \right)AXB+\alpha \left(\frac{{A}^{2}X+X{B}^{2}}{2}\right)\parallel +2\left(2u-1\right)\parallel AXB\parallel .$
So,
$\begin{array}{rcl}\parallel {A}^{u}X{B}^{2-u}+{A}^{2-u}X{B}^{u}\parallel & \le & 2\left(4{r}_{0}-1\right)\parallel AXB\parallel \\ +4\left(1-2{r}_{0}\right)\parallel \left(1-\alpha \right)AXB+\alpha \left(\frac{{A}^{2}X+X{B}^{2}}{2}\right)\parallel .\end{array}$
(2.3)
For $1\le u\le \frac{3}{2}$, by the convexity of the function $g\left(u\right)$ and Lemma 2.1, presented above, we have
$g\left(u\right)\le \frac{g\left(\frac{3}{2}\right)-g\left(1\right)}{\frac{1}{2}}u-\frac{g\left(\frac{3}{2}\right)-\frac{3}{2}g\left(1\right)}{\frac{1}{2}},$
which implies
$g\left(u\right)\le \left(3-2u\right)g\left(1\right)+2\left(u-1\right)g\left(\frac{3}{2}\right).$
(2.4)
By (1.3) and (2.4), we have
$\begin{array}{rcl}\parallel {A}^{u}X{B}^{2-u}+{A}^{2-u}X{B}^{u}\parallel & \le & 2\left(3-2u\right)\parallel AXB\parallel \\ +4\left(u-1\right)\parallel \left(1-\alpha \right)AXB+\alpha \left(\frac{{A}^{2}X+X{B}^{2}}{2}\right)\parallel .\end{array}$
So,
$\begin{array}{rcl}\parallel {A}^{u}X{B}^{2-u}+{A}^{2-u}X{B}^{u}\parallel & \le & 2\left(4{r}_{0}-1\right)\parallel AXB\parallel \\ +4\left(1-2{r}_{0}\right)\parallel \left(1-\alpha \right)AXB+\alpha \left(\frac{{A}^{2}X+X{B}^{2}}{2}\right)\parallel .\end{array}$
(2.5)
By (2.3) and (2.5), for $\frac{1}{2}\le u\le \frac{3}{2}$, $\alpha \in \left[\frac{1}{2},\mathrm{\infty }\right)$ and ${r}_{0}=min\left[\frac{u}{2},1-\frac{u}{2}\right]$, we have the following equivalent inequality:
$\begin{array}{rcl}\parallel {A}^{u}X{B}^{2-u}+{A}^{2-u}X{B}^{u}\parallel & \le & 2\left(4{r}_{0}-1\right)\parallel AXB\parallel \\ +4\left(1-2{r}_{0}\right)\parallel \left(1-\alpha \right)AXB+\alpha \left(\frac{{A}^{2}X+X{B}^{2}}{2}\right)\parallel .\end{array}$

The proof is completed. □

Remark 2.3 With a simple computation between the upper bounds in (1.3) and (2.1), obviously we have
$\begin{array}{c}\parallel \left(1-\alpha \right)AXB+\alpha \left(\frac{{A}^{2}X+X{B}^{2}}{2}\right)\parallel \hfill \\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}-\left(4{r}_{0}-1\right)\parallel AXB\parallel -2\left(1-2{r}_{0}\right)\parallel \left(1-\alpha \right)AXB+\alpha \left(\frac{{A}^{2}X+X{B}^{2}}{2}\right)\parallel \hfill \\ \phantom{\rule{1em}{0ex}}=\left(4{r}_{0}-1\right)\parallel \left(1-\alpha \right)AXB+\alpha \left(\frac{{A}^{2}X+X{B}^{2}}{2}\right)\parallel -\left(4{r}_{0}-1\right)\parallel AXB\parallel \hfill \\ \phantom{\rule{1em}{0ex}}=\left(4{r}_{0}-1\right)\left(\parallel \left(1-\alpha \right)AXB+\alpha \left(\frac{{A}^{2}X+X{B}^{2}}{2}\right)\parallel -\parallel AXB\parallel \right)\hfill \\ \phantom{\rule{1em}{0ex}}>0.\hfill \end{array}$

Thus the inequality (2.1) is a refinement of the inequality (1.3).

Now, we present a refinement of the inequality $\parallel AXB\parallel \le \parallel \left(1-\alpha \right)AXB+\alpha \left(\frac{{A}^{2}X+X{B}^{2}}{2}\right)\parallel$.

Theorem 2.4 Let $A,B,X\in {M}_{n}$, such that A, B are positive semidefinite. Then for any unitarily invariant norm $\parallel \cdot \parallel$, $\frac{1}{2}\le u\le \frac{3}{2}$, and $\alpha \in \left[\frac{1}{2},\mathrm{\infty }\right)$, we have
$\begin{array}{c}2\parallel AXB\parallel +2\left({\int }_{\frac{1}{2}}^{\frac{3}{2}}\parallel {A}^{u}X{B}^{2-u}+{A}^{2-u}X{B}^{u}\parallel \phantom{\rule{0.2em}{0ex}}du-2\parallel AXB\parallel \right)\hfill \\ \phantom{\rule{1em}{0ex}}\le \parallel \left(1-\alpha \right)AXB+\alpha \left(\frac{{A}^{2}X+X{B}^{2}}{2}\right)\parallel ,\hfill \end{array}$
(2.6)

where ${r}_{0}=min\left[\frac{u}{2},1-\frac{u}{2}\right]$.

Proof For $\frac{1}{2}\le u\le 1$, from Theorem 2.2, we have
$\begin{array}{rcl}\parallel {A}^{u}X{B}^{2-u}+{A}^{2-u}X{B}^{u}\parallel & \le & 4\left(1-u\right)\parallel \left(1-\alpha \right)AXB+\alpha \left(\frac{{A}^{2}X+X{B}^{2}}{2}\right)\parallel \\ +2\left(2u-1\right)\parallel AXB\parallel .\end{array}$
By integrating both sides of the inequality above, we have
$\begin{array}{c}{\int }_{\frac{1}{2}}^{1}\parallel {A}^{u}X{B}^{2-u}+{A}^{2-u}X{B}^{u}\parallel \phantom{\rule{0.2em}{0ex}}du\hfill \\ \phantom{\rule{1em}{0ex}}\le 4\parallel \left(1-\alpha \right)AXB+\alpha \left(\frac{{A}^{2}X+X{B}^{2}}{2}\right)\parallel {\int }_{\frac{1}{2}}^{1}\left(1-u\right)\phantom{\rule{0.2em}{0ex}}du\hfill \\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}+2\parallel AXB\parallel {\int }_{\frac{1}{2}}^{1}\left(2u-1\right)\phantom{\rule{0.2em}{0ex}}du\hfill \\ \phantom{\rule{1em}{0ex}}=\frac{1}{2}\parallel \left(1-\alpha \right)AXB+\alpha \left(\frac{{A}^{2}X+X{B}^{2}}{2}\right)\parallel +\frac{1}{2}\parallel AXB\parallel .\hfill \end{array}$
(2.7)
For $1\le u\le \frac{3}{2}$, from Theorem 2.2, we have
$\begin{array}{rcl}\parallel {A}^{u}X{B}^{2-u}+{A}^{2-u}X{B}^{u}\parallel & \le & 2\left(3-2u\right)\parallel AXB\parallel \\ +4\left(u-1\right)\parallel \left(1-\alpha \right)AXB+\alpha \left(\frac{{A}^{2}X+X{B}^{2}}{2}\right)\parallel .\end{array}$
Similarly, by integrating both sides of the inequality above, we have
$\begin{array}{c}{\int }_{1}^{\frac{3}{2}}\parallel {A}^{u}X{B}^{2-u}+{A}^{2-u}X{B}^{u}\parallel \phantom{\rule{0.2em}{0ex}}du\hfill \\ \phantom{\rule{1em}{0ex}}\le 2\parallel AXB\parallel {\int }_{1}^{\frac{3}{2}}\left(3-2u\right)\phantom{\rule{0.2em}{0ex}}du\hfill \\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}+4\parallel \left(1-\alpha \right)AXB+\alpha \left(\frac{{A}^{2}X+X{B}^{2}}{2}\right)\parallel {\int }_{1}^{\frac{3}{2}}\left(2u-1\right)\phantom{\rule{0.2em}{0ex}}du\hfill \\ \phantom{\rule{1em}{0ex}}=\frac{1}{2}\parallel \left(1-\alpha \right)AXB+\alpha \left(\frac{{A}^{2}X+X{B}^{2}}{2}\right)\parallel +\frac{1}{2}\parallel AXB\parallel .\hfill \end{array}$
(2.8)
It follows from (2.7) and (2.8) that
${\int }_{\frac{1}{2}}^{\frac{3}{2}}\parallel {A}^{u}X{B}^{2-u}+{A}^{2-u}X{B}^{u}\parallel \phantom{\rule{0.2em}{0ex}}du\le \parallel \left(1-\alpha \right)AXB+\alpha \left(\frac{{A}^{2}X+X{B}^{2}}{2}\right)\parallel +\parallel AXB\parallel ,$
which is equivalent to
$\begin{array}{c}2\parallel AXB\parallel +2\left({\int }_{\frac{1}{2}}^{\frac{3}{2}}\parallel {A}^{u}X{B}^{2-u}+{A}^{2-u}X{B}^{u}\parallel \phantom{\rule{0.2em}{0ex}}du-2\parallel AXB\parallel \right)\hfill \\ \phantom{\rule{1em}{0ex}}\le 2\parallel \left(1-\alpha \right)AXB+\alpha \left(\frac{{A}^{2}X+X{B}^{2}}{2}\right)\parallel .\hfill \end{array}$

The proof is completed. □

Remark 2.5 Obviously,
${\int }_{\frac{1}{2}}^{\frac{3}{2}}\parallel {A}^{u}X{B}^{2-u}+{A}^{2-u}X{B}^{u}\parallel \phantom{\rule{0.2em}{0ex}}du-2\parallel AXB\parallel \ge 0.$

Thus, the inequality (2.6) is a refinement of the inequality $\parallel AXB\parallel \le \parallel \left(1-\alpha \right)AXB+\alpha \left(\frac{{A}^{2}X+X{B}^{2}}{2}\right)\parallel$.

Taking $\alpha =\frac{2}{t+2}$ ($-2), the following corollaries are obtained.

Corollary 2.6 Let $A,B,X\in {M}_{n}$, such that A, B are positive semidefinite. Then for any unitarily invariant norm $\parallel \cdot \parallel$ and $\frac{1}{2}\le u\le \frac{3}{2}$,
$\begin{array}{rcl}\parallel {A}^{u}X{B}^{2-u}+{A}^{2-u}X{B}^{u}\parallel & \le & 2\left(4{r}_{0}-1\right)\parallel AXB\parallel \\ +\frac{4\left(1-2{r}_{0}\right)}{t+2}\parallel {A}^{2}X+tAXB+X{B}^{2}\parallel ,\end{array}$
(2.9)

where ${r}_{0}=min\left[\frac{u}{2},1-\frac{u}{2}\right]$ and $-2.

Corollary 2.7 Let $A,B,X\in {M}_{n}$, such that A, B are positive semidefinite. Then for any unitarily invariant norm $\parallel \cdot \parallel$, $\frac{1}{2}\le u\le \frac{3}{2}$,
$\begin{array}{c}2\parallel AXB\parallel +2\left({\int }_{\frac{1}{2}}^{\frac{3}{2}}\parallel {A}^{u}X{B}^{2-u}+{A}^{2-u}X{B}^{u}\parallel \phantom{\rule{0.2em}{0ex}}du-2\parallel AXB\parallel \right)\hfill \\ \phantom{\rule{1em}{0ex}}\le \frac{2}{t+2}\parallel {A}^{2}X+tAXB+X{B}^{2}\parallel ,\hfill \end{array}$
(2.10)

where ${r}_{0}=min\left[\frac{u}{2},1-\frac{u}{2}\right]$ and $-2.

Thus, on the one hand, the inequality (2.9) is a refinement of the inequality $\parallel AXB\parallel \le \frac{1}{t+2}\parallel {A}^{2}X+tAXB+X{B}^{2}\parallel$, and also another version of the inequality (1.6); on the other hand, the inequality (2.10) is just the inequality proved in , so the inequality (2.6) presented in Theorem 2.4 is also the generalization of the inequality proved in .

## Declarations

### Acknowledgements

The authors wish to express their heartfelt thanks to the referees for their detailed and helpful suggestions for revising the manuscript.

## Authors’ Affiliations

(1)
College of Mathematics and Statistics, Chongqing University, Chongqing, 401331, P.R., China

## References 